<p style='text-indent:20px;'>Let <inline-formula><tex-math id="M2">\begin{document}$ T_{a, \varphi} $\end{document}</tex-math></inline-formula> be a Fourier integral operator defined by the oscillatory integral</p><p style='text-indent:20px;'><disp-formula> <label/> <tex-math id="FE1"> \begin{document}$ \begin{eqnarray*} T_{a, \varphi}u(x) & = &\frac{1}{(2\pi)^n}\int_{\mathbb{R}^n} e^{ i \varphi(x, \xi)}a(x, \xi) \hat{u}(\xi)d\xi, \end{eqnarray*} $\end{document} </tex-math></disp-formula></p><p style='text-indent:20px;'>where <inline-formula><tex-math id="M3">\begin{document}$ a\in L^{p} S^{m}_{\varrho} $\end{document}</tex-math></inline-formula> (<inline-formula><tex-math id="M4">\begin{document}$ 1\leq p\leq\infty, $\end{document}</tex-math></inline-formula> <inline-formula><tex-math id="M5">\begin{document}$ 0\leq\varrho\leq1 $\end{document}</tex-math></inline-formula>) and <inline-formula><tex-math id="M6">\begin{document}$ \varphi\in L^{\infty}\Phi^{2} $\end{document}</tex-math></inline-formula>, satisfying the rough non-degenerate condition. It is showed that if <inline-formula><tex-math id="M7">\begin{document}$ 0<r\leq \infty, $\end{document}</tex-math></inline-formula> <inline-formula><tex-math id="M8">\begin{document}$ 1\leq q\leq\infty $\end{document}</tex-math></inline-formula> satisfying the relation <inline-formula><tex-math id="M9">\begin{document}$ \frac{1}{r} = \frac{1}{q}+\frac{1}{p} $\end{document}</tex-math></inline-formula>, then <inline-formula><tex-math id="M10">\begin{document}$ T_{a, \varphi} $\end{document}</tex-math></inline-formula> is a bounded operator from <inline-formula><tex-math id="M11">\begin{document}$ L^{q}(\mathbb{R}^n) $\end{document}</tex-math></inline-formula> to <inline-formula><tex-math id="M12">\begin{document}$ L^{r}(\mathbb{R}^n) $\end{document}</tex-math></inline-formula> provided</p><p style='text-indent:20px;'><disp-formula> <label/> <tex-math id="FE2"> \begin{document}$ m<-\varrho\frac{(n-1)}{2}\big(\frac{1}{s}+\frac{1}{\min(p, s')}\big)+\frac{n(\varrho-1)}{s}, $\end{document} </tex-math></disp-formula></p><p style='text-indent:20px;'>where <inline-formula><tex-math id="M13">\begin{document}$ s = \min(2, p, q) $\end{document}</tex-math></inline-formula> and <inline-formula><tex-math id="M14">\begin{document}$ \frac{1}{s}+\frac{1}{s'} = 1. $\end{document}</tex-math></inline-formula></p>